Problem: You are given N items.
The value of the i-th item (1 \leq i \leq N) is v_i.
Your have to select at least A and at most B of these items.
Under this condition, find the maximum possible arithmetic mean of the values of selected items.
Additionally, find the number of ways to select items so that the mean of the values of selected items is maximized.

Constraints

* 1 \leq N \leq 50
* 1 \leq A,B \leq N
* 1 \leq v_i \leq 10^{15}
* Each v_i is an integer.

Input

The input is given from Standard Input in the following format:


N A B
v_1
v_2
...
v_N


Output

Print two lines.
The first line should contain the maximum possible arithmetic mean of the values of selected items. The output should be considered correct if the absolute or relative error is at most 10^{-6}.
The second line should contain the number of ways to select items so that the mean of the values of selected items is maximized.

Examples

Input

5 2 2
1 2 3 4 5


Output

4.500000
1


Input

4 2 3
10 20 10 10


Output

15.000000
3


Input

5 1 5
1000000000000000 999999999999999 999999999999998 999999999999997 999999999999996


Output

1000000000000000.000000
1

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		// 要素の入力、格納
		int N = sc.nextInt();
		int A = sc.nextInt();
		int B = sc.nextInt();
		double v[] = new double[N];
		for (int i = 0; i < N; i++) {
			v[i] = sc.nextLong();
		}
		Arrays.sort(v);

		// パスカル三角形の準備
		long pascal[][] = Pascal(N);

		// A個の際の平均値の計算
		double suma = 0;
		for (int i = 0; i < A; i++) {
			suma = suma + v[N - 1 - i];
		}
		double avea = (double) suma / A;
		String ansave = String.format("%.6f", avea);

		// A個の場合の組み合わせ
		// A以上の中にAと同値がいくつあるか
		int samea = 1; // 自分自身
		for (int i = 0; i < A - 1; i++) {
			if (v[N - A] == v[N - A + 1 + i]) {
				samea += 1;
			} else {
				break;
			}
		}
		int bunboa = samea;
		// A以下の中にAと同値がいくつあるか
		for (int i = 0; i < N - A; i++) {
			if (v[N - A] == v[N - A - i - 1]) {
				samea += 1;
			} else {
				break;
			}
		}
		int bunshia = samea;
		long comba = pascal[bunshia][bunboa]; // bunshia C bunboa の計算

		// A個以上の場合
		int sameb = 0;
		long combb = 0;
		long sumcombb = 0;
		if (B > A) {
			// 最大値と同値がいくつあるか
			for (int i = 0; i < N; i++) {
				if (v[N - 1 - i] == v[N - 1]) {
					sameb += 1;
				}
			}
			// A+1〜Bについてそれぞれの組み合わせ数を累計
			for (int i = 0; i < B - A; i++) {
				combb = pascal[sameb][A + 1 + i];
				sumcombb += combb;
			}
		}

		long sumcombab = comba + sumcombb;
		// 出力
		System.out.println(ansave);
		System.out.println(sumcombab);
	}

	// パスカル三角形導出メソッド
	public static long[][] Pascal(int max) {
		long pascal[][] = new long[max + 1][max + 1];
		pascal[0][0] = 1;
		for (int i = 0; i < max + 1; i++) {
			for (int j = 0; j < i + 1; j++) {
				if (i == 0 && j == 0) {
					continue;
				} else if (j == 0) {
					pascal[i][j] = pascal[i - 1][j];
				} else if (j == i) {
					pascal[i][j] = pascal[i - 1][j - 1];
				} else {
					pascal[i][j] = pascal[i - 1][j - 1] + pascal[i - 1][j];
				}
			}
		}
		return pascal;
	}
}